11. 计算：
(1) $\dfrac{3}{5}-3.7-(-\dfrac{2}{5})-1.3$；
(2) $7-(+8)-(-2)$；
(3) $(+7.2)-(+3.6)-(-3.6)-(-2.8)$；
(4) $\dfrac{2}{5}-\left|-1\dfrac{1}{2}\right|-(+2\dfrac{1}{4})-(-2.75)$.

11.解:(1)原式$=\dfrac{3}{5}+\dfrac{2}{5}+[(-3.7)+(-1.3)]=1+(-5)=-4$.
(2)原式$=7-8+2=7+2-8=1$.
(3)原式$=7.2-3.6+3.6+2.8=7.2+2.8+3.6-3.6=10$.
(4)原式$=\dfrac{2}{5}-1\dfrac{1}{2}-2\dfrac{1}{4}+2.75=0.4-1.5-2.25+2.75=(0.4-1.5)+(-2.25+2.75)=-1.1+0.5=-0.6$.

12. 某中学图书馆某一周借书记录如下:(以100册为基准,超过的册数记为正,不足的册数记为负)

(1)星期五借出多少册书?
(2)星期四比星期三多借出多少册书?
(3)该周平均每天借出多少册书?

12.解:(1)$100+(-12)=88$(册).
答:星期五借出88册书.
(2)$[100+(+6)]-[100+(-17)]=23$(册).
答:星期四比星期三多借出23册书.
(3)$100+[(+23)+0+(-17)+(+6)+(-12)]÷5=100$(册).
答:该周平均每天借出100册书.

13. 若$\left|\dfrac{1}{2}-1\right|=1-\dfrac{1}{2}$，$\left|\dfrac{1}{3}-\dfrac{1}{2}\right|=\dfrac{1}{2}-\dfrac{1}{3}$，$\left|\dfrac{1}{4}-\dfrac{1}{3}\right|=\dfrac{1}{3}-\dfrac{1}{4}$，…，探索规律，并解答问题.
(1)$\left|\dfrac{1}{10}-\dfrac{1}{9}\right|=$；
(2)计算：$\left|\dfrac{1}{2}-1\right|+\left|\dfrac{1}{3}-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}-\dfrac{1}{3}\right|+\left|\dfrac{1}{5}-\dfrac{1}{4}\right|=$；
(3)计算：$\left|\dfrac{1}{2}-1\right|+\left|\dfrac{1}{3}-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}-\dfrac{1}{3}\right|+\dots +\left|\dfrac{1}{2026}-\dfrac{1}{2025}\right|$.

13.(1)$\dfrac{1}{9}-\dfrac{1}{10}$
(2)$\dfrac{4}{5}$
(3)解:原式$=(1-\dfrac{1}{2})+(\dfrac{1}{2}-\dfrac{1}{3})+(\dfrac{1}{3}-\dfrac{1}{4})+\dots+(\dfrac{1}{2025}-\dfrac{1}{2026})=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dots+\dfrac{1}{2025}-\dfrac{1}{2026}=1-\dfrac{1}{2026}=\dfrac{2025}{2026}$.

14. (2024·吴中区月考)已知$|5-2|$表示5与2两个数在数轴上所对应的两个点之间的距离.
(1)求$|5-(-2)|$的值;
(2)如果$|x+2|=1$,请写出$x$的值;
(3)求适合条件$|x-1|＜3$的所有整数$x$的值.

14.解:(1)$|5-(-2)|=7$.
(2)因为$|x+2|=1$,
所以$x+2=\pm1$,解得$x=-3$或$x=-1$.
(3)$|x-1|＜3$表示$x$与1两个数在数轴上所对应的两个点之间的距离小于3,所以整数$x$的值有$-1,0,1,2,3$.